HOT Competition: Is |x+y| |x-y| ? (1) |x| |y| (2) |x - y|

Question

Is  |x+y| > |x-y|  ?

(1)  |x| > |y|

(2)  |x - y| < |x|

Bunuel · Accepted Answer

Is |x + y| > |x - y| ? Is |x+y| > |x-y| ? --> square it: is x^2+2xy+y^2>x^2-2xy+y^2 --> is xy>0 ? (1) |x| > |y|. We cannot get from this whether x and y have the same sign. Not sufficient. (2) |x - y| < |x| --> square again: x^2-2xy+y^2 xy>\frac{y^2}{2} --> since y^2\geq{0} (the square of any number is more than or equal to 0), then we have that xy>\frac{y^2}{2}\geq{0} . Sufficient. Answer: B.

yashikaaggarwal · Answer

Statement 1: |X| > |Y|
X>±Y
Case 1: X>Y
X-Y > 0

Case 2: X>-Y
X+Y > 0

No sufficient constraint to determine answer.
(Insufficient)

Statement 2: |x-y|<|X|
If |x-y|<|X|
|x-y| will definitely be less than |X+Y|
(Sufficient)

Answer is B

Posted from my mobile device

GMATinsight · Answer

Answer: Option B

Please check the video for the step-by-step solution.

GMATinsight's Solution

https://www.youtube.com/watch?v=dzm4OJrmGjQ

Archit3110 · Answer

Target
Is |x+y|>|x−y| ?

#1
|x|>|y|
check at x ,y pairs of ( 2,1) & (-2,-1) we get yes and
at x,y pairs of (-1,1/2) we get no to our target ; insufficient
#2
|x−y|<|x|
possible at x,y pairs of same sign eg ( -2,-1 ) & ( 2,1)
we get yes for our target value ; sufficient
OPTION B

Is |x+y|>|x−y| ?

(1) |x|>|y||x|>|y|

(2) |x−y|<|x|

ShankSouljaBoi · Answer

Squaring the stem
---> 2xy > -2xy
===> 4xy >0 --- rephrasing is xy >0 or do x and y have opposite signs ?

1. Absolute values are given. We can't comment on polarity of the variables.  Insufficient 
2. Squaring --> y^2 - 2x < 0 ---> 2x > y^2 ----> minnimum value of Y^2 can be zero .... SO R.H.S. is at least Zero
Hence, LHS > 0 .... 2x > 0 ---> x > 0 . But we dont know the polarity of Y.  Insufficient

Combining ,

We know that absolute value of X is more than the absolute value of Y
and x>0

These are still  not suffcient   as we dont know the polairty of Y. For example,
consider x = 3 and y =-1/2 . Our answer is NO
and x = 3 and y =+1/2 . Our answer is Yes.
Hence,  E

Ranasaymon · Answer

Is |x+y|>|x−y| ?

(St 1) |x|>|y|
If, |2|>|1|; then |2+1|>|2−1|; |3|>|1|? Yes
If, |2|>|-1|; then |2-1|>|2+1|; |1|>|3|? No

St. 1 has more than 1 value, so it is insufficient

(St 2) |x−y|<|x|
If x=-2 and y=-1; then |-2-1|>|-2+1|; |-3|>|-1|? Yes
If x=2 and y=1; then |2+1|>|2−1|; |3|>|1|? Yes

St. 2 is sufficient as it has one value

ANS. B

acedrmonstaa · Answer

|x−y|>|x+y| square both sides to get

x2+y2−2xy>x2+y2+2xy=>4xy<0

so the question Is xy<0 (means are they of opposite signs)

1) |x|>|y| --> x can be 2 or y can be 1 (same sign) or x can be 2 and y can be -1. (Insufficient)
2) |x-y|>|y| --> using the above example, if x=2, y=1, the statement is null. However, if x and y are opposite signs, the statement is always true. Plug in another example (x = -2, y = 1). Using same sign x=-2 and y=-3 also does not work. (Sufficient)

sood1596 · Answer

Is |x+y|>|x−y| ?

(1) |x|>|y|
Case1: x=2; y=1
3>1 Yes
Case2: x=-2 y=1
1>3 No

Cancel AD

(2) |x−y|<|x|
Here notice we cant take a negative value to test because this equation wont be true, so
x=2 y =1
Yes it is sufficient to answer the question

Answer - B

RaghavKhanna · Answer

Is |x+y| > |x-y| ?

Squaring both sides:
Is (x+y)^2 > (x-y)^2
Is xy > 0 ?

Statement 1:
|x| > |y|
(i) x = 10, y = 5
=> xy > 0
(ii) x = 10, y = -5
=> xy < 0

Statement 1 is insufficient

Statement 2:
|x-y| < |x|
=> (x-y)^2 < x^2
=> y^2 - 2xy < 0
=> y(y-2x) < 0

Either y < 0 and y - 2x > 0
Or
y - 2x < 0 and y > 0

In either case, xy > 0

Statement 2 is sufficient

Answer: B

Posted from my mobile device

GMATGuruNY · Answer

Number line approach:

|x-y| = the distance between x and y
|x+y| = |x - (-y)| = the distance between x and -y

Is |x-y| < |x+y| ?
In words:
 Is the distance between x and y less than the distance between x and -y?

..........-y.......... 0 ..........y.......... 
 ...........y.......... 0 ..........-y..........  
In each number line:
For x to be closer to y than to -y, x and y must be together in a green area.
In other words, x and y must have the SAME SIGN.

Question stem, rephrased: 
Do x and y have the same sign?

Statement 1: 
No way to determine whether x and y have the same sign.
INSUFFICIENT.

Statement 2, put into words: 
The distance between x and y is less than the distance between x and 0.
 ........................0.....| .....y.......... 
 ...........y..... |.....0........................ 
In each number line:
The vertical line lies halfway between y and 0.
For x to be closer to y than to 0, x and y must be together in a green area.
Implication:
x and y have the SAME SIGN.
Thus, the answer to the rephrased question stem is YES.
SUFFICIENT.

B

exc4libur · Answer

|x+y|>|x−y| only true if x and y are same sign

(1) insufic
x,y could be any sign

(2) sufic
only true if x and y are different signs
x,y=1,-1: 1--1=|2|>|1|
x,y=-2,2: -2-2=|-4|>|-2|

(B)

MakSha · Answer

Hi  Bunuel ,

Great explanation. Just one query - what if x = y = 0? Since the same is not stated explicitly in the question, is the equality implied?

Thanks

Bunuel · Answer

This case is not possible because x = y = 0 does not satisfy |x - y| < |x|.

Thekingmaker · Answer

it's a quite simple question A being clearly insuff i am gonna just concentrate on B
B it states that y>0 is y was negative then it would have added while subtracting and would have given a different result 
imo B

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

HOT Competition: Is |x+y| > |x-y| ? (1) |x| > |y| (2) |x - y| < |x|

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions